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I'm not very confident with Galois Theory, so sorry if these are basics questions and there are 5 questions. $K$ is an algebraic cubic number field, and we fix an algebraic closure of rationals $\overline{\mathbb{Q}}$. With $ K^{Gal}$ I denote the Galois closure of $ K $. With $ \operatorname{Hom}_{\mathbb{Q}}(K,\overline{\mathbb{Q}}) $ the set of embeddings, and $ \sigma $ it is always an embendding $ \sigma \in \operatorname{Hom}_{\mathbb{Q}}(K,\overline{\mathbb{Q}})$.

Q1. Clearly if $ \sigma \in \operatorname{Hom}_{\mathbb{Q}}(K,\overline{\mathbb{Q}}) $ then $ \sigma(K) $ and $K$ are conjugates fields (isomorphic). It is possible that there is a field $K'$ isomorphic to $K$ but for which it is not equal to $ \sigma(K)$ for any $ \sigma \in \operatorname{Hom}_{\mathbb{Q}}(K,\overline{\mathbb{Q}}) $ ??

Q2. I don't get why $ [ K^{Gal} : \mathbb{Q} ]=6$ with Galois group hence that is isomorphic to $S_3$.

Q3. If $K= \mathbb{Q}(\theta)$ it is true that $ K= \mathbb{Q}(\sigma(\theta)) $ (equal not just isomorphic) when $K/\mathbb{Q} $ is a Galois extension, if yes why?

Q4. If $K= \mathbb{Q}(\theta)$ but the $K/\mathbb{Q}$ is not a Galois extension then it is true that $ K $ is isomorphic to $ \mathbb{Q}(\sigma(\theta)) $?

Q5. The element of $ K^{Gal}$ that are invariant under the Galois group of $ K^{Gal} $ are only the element of $\mathbb{Q} $ ?

Q6. These question (except Q2. of course) have same answer if $K$ is not a cubic number field but an arbitrarly number field?

3m0o
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    Yes, I think it is better not to ask so many questions in one post. If you do some more examples, some questions will automatically dissolve, too. – Dietrich Burde Dec 25 '21 at 09:19
  • For Q1 an isomorphism $\sigma:K\to K'\subset \overline{\Bbb{Q}}$ is automatically an element of $Hom_\Bbb{Q}(K,\overline{\Bbb{Q}})$. – reuns Dec 25 '21 at 11:51

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Q1: If $K \cong \sigma(K)$, this isomorphism must give you a map $K \to \sigma(K) \subset \overline{\mathbb{Q}}$ automatically

Q2: That is not true. You can see a degree $3$ Galois extension here

If what you mean is that if the galois closure has degree $6$, then the Galois group is $S_3$, then that is true. This is because if $K = \mathbb{Q}(\theta)$ (which is possible by the theorem of the primitive element), then the Galois closure is the splitting extension of the minimal polynomial of $\theta$, which must be of degree $3$ because it is a cubic extension. But then the Galois group acts on the roots by permuting them (and any automorphism of the field is determined by the action on these roots so they generate the field) so it must be a subgroup of $S_3$.

Finally, $\text{Gal}(K^\text{Gal}| \mathbb{Q}) = [K^\text{Gal}: \mathbb{Q}]$ and $S_3$ is its only subgroup with $6$ elements.

Q3: Yes, this is true. Normality gives you that $\sigma (\theta) \in K$, since $\theta$ must be a root of the minimal polynomial of $\theta$ (roots are preserved by $\mathbb{Q}$-automorphisms). Then, $\deg m_\theta (x) = [K : \mathbb{Q}] = [\mathbb{Q}(\sigma(\theta)): \mathbb{Q}]$. Since it is a subextension and it has the same degree it must be equal.

Q4: Yes, they are both isomorphism to $\mathbb{Q}[x]/m_\theta(x)$.

Q5: Yes, this is an easy consequence of the fundamental theorem of Galois theory.

Q6: Yes

Sofía Marlasca Aparicio
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  • Thank you very much marlasca23, for Q2. i meant of course when $K/\mathbb{Q}$ is not a Galois extension then $ [K^{Gal} : \mathbb{Q}]=6$. – 3m0o Dec 25 '21 at 09:58
  • I think i got it, $ [ K^{Gal} : \mathbb{Q} ] \mid 6 $ and $ [ K^{Gal} : \mathbb{Q} ] =[K^{Gal} : K ][K : \mathbb{Q}] = 3[K^{Gal} : K ]$, since $ K^{Gal}/\mathbb{Q}$ is Galois and $ K/\mathbb{Q}$ is not we conclude that $[K^{Gal} : K ]=2$ – 3m0o Dec 26 '21 at 00:43